On showing that $varphi(N)=(N+K)/K$, for modules $N,K<M$, $varphi$ natural map.
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In this question modules are A-modules, where A is a unitary commutative ring
This seems trivial, but beginning with module theory, I was trying to see why $varphi(N)=(N+K)/K$, for the natural A-module homomorphism $varphi:Mto M/K$, and submodules $N,Ksubset M$.
In doing so, I got a bit confused over why we don't just have $(N+K)/K=N/K$. My thoughts are these:
$N+K={n+k:nin N, kin K}$, and $(N+K)/K={(n+k)+K:(n+k)in (N+K)}$, $N/K={n+K:nin N}$, as sets.
And isn't $(n+k)+K=n+K$, for some $nin N, kin K$?
And then, for the quotients we have equivalence relations, so that
$N/K nibar{n_1}=bar{n_2}Leftrightarrow n_1 + K=n_2+KLeftrightarrow (n_1-n_2)in K$.
Well, we also have
$(N+K)/K nioverline{n_1+k_1}=overline{n_2+k_2}Leftrightarrow n_1 + k_1 + K=n_2+k_2 +KLeftrightarrow (n_1-n_2)+(k_1-k_2)in K Leftrightarrow (n_1-n_2)in K$.
Thus, intuitively at least, it seems we should have $(N+K)/K=N/K$.
Now,
Question1: Where have I gone wrong, misunderstood or forgotten something?
Question2: Will someone provide a proof sketch of the statement in the title?
abstract-algebra ring-theory modules
edited Nov 22 at 21:34
user26857
39.2k123882
asked Nov 22 at 15:46
Christopher.L
7571317
add a comment |
up vote
1
down vote
favorite
In this question modules are A-modules, where A is a unitary commutative ring
This seems trivial, but beginning with module theory, I was trying to see why $varphi(N)=(N+K)/K$, for the natural A-module homomorphism $varphi:Mto M/K$, and submodules $N,Ksubset M$.
In doing so, I got a bit confused over why we don't just have $(N+K)/K=N/K$. My thoughts are these:
$N+K={n+k:nin N, kin K}$, and $(N+K)/K={(n+k)+K:(n+k)in (N+K)}$, $N/K={n+K:nin N}$, as sets.
And isn't $(n+k)+K=n+K$, for some $nin N, kin K$?
And then, for the quotients we have equivalence relations, so that
$N/K nibar{n_1}=bar{n_2}Leftrightarrow n_1 + K=n_2+KLeftrightarrow (n_1-n_2)in K$.
Well, we also have
$(N+K)/K nioverline{n_1+k_1}=overline{n_2+k_2}Leftrightarrow n_1 + k_1 + K=n_2+k_2 +KLeftrightarrow (n_1-n_2)+(k_1-k_2)in K Leftrightarrow (n_1-n_2)in K$.
Thus, intuitively at least, it seems we should have $(N+K)/K=N/K$.
Now,
Question1: Where have I gone wrong, misunderstood or forgotten something?
Question2: Will someone provide a proof sketch of the statement in the title?
abstract-algebra ring-theory modules
edited Nov 22 at 21:34
user26857
39.2k123882
asked Nov 22 at 15:46
Christopher.L
7571317
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In this question modules are A-modules, where A is a unitary commutative ring
This seems trivial, but beginning with module theory, I was trying to see why $varphi(N)=(N+K)/K$, for the natural A-module homomorphism $varphi:Mto M/K$, and submodules $N,Ksubset M$.
In doing so, I got a bit confused over why we don't just have $(N+K)/K=N/K$. My thoughts are these:
$N+K={n+k:nin N, kin K}$, and $(N+K)/K={(n+k)+K:(n+k)in (N+K)}$, $N/K={n+K:nin N}$, as sets.
And isn't $(n+k)+K=n+K$, for some $nin N, kin K$?
And then, for the quotients we have equivalence relations, so that
$N/K nibar{n_1}=bar{n_2}Leftrightarrow n_1 + K=n_2+KLeftrightarrow (n_1-n_2)in K$.
Well, we also have
$(N+K)/K nioverline{n_1+k_1}=overline{n_2+k_2}Leftrightarrow n_1 + k_1 + K=n_2+k_2 +KLeftrightarrow (n_1-n_2)+(k_1-k_2)in K Leftrightarrow (n_1-n_2)in K$.
Thus, intuitively at least, it seems we should have $(N+K)/K=N/K$.
Now,
Question1: Where have I gone wrong, misunderstood or forgotten something?
Question2: Will someone provide a proof sketch of the statement in the title?
abstract-algebra ring-theory modules
edited Nov 22 at 21:34
user26857
39.2k123882
asked Nov 22 at 15:46
Christopher.L
7571317
In this question modules are A-modules, where A is a unitary commutative ring
This seems trivial, but beginning with module theory, I was trying to see why $varphi(N)=(N+K)/K$, for the natural A-module homomorphism $varphi:Mto M/K$, and submodules $N,Ksubset M$.
In doing so, I got a bit confused over why we don't just have $(N+K)/K=N/K$. My thoughts are these:
$N+K={n+k:nin N, kin K}$, and $(N+K)/K={(n+k)+K:(n+k)in (N+K)}$, $N/K={n+K:nin N}$, as sets.
And isn't $(n+k)+K=n+K$, for some $nin N, kin K$?
And then, for the quotients we have equivalence relations, so that
$N/K nibar{n_1}=bar{n_2}Leftrightarrow n_1 + K=n_2+KLeftrightarrow (n_1-n_2)in K$.
Well, we also have
$(N+K)/K nioverline{n_1+k_1}=overline{n_2+k_2}Leftrightarrow n_1 + k_1 + K=n_2+k_2 +KLeftrightarrow (n_1-n_2)+(k_1-k_2)in K Leftrightarrow (n_1-n_2)in K$.
Thus, intuitively at least, it seems we should have $(N+K)/K=N/K$.
Now,
Question1: Where have I gone wrong, misunderstood or forgotten something?
Question2: Will someone provide a proof sketch of the statement in the title?
abstract-algebra ring-theory modules
abstract-algebra ring-theory modules
edited Nov 22 at 21:34
user26857
39.2k123882
asked Nov 22 at 15:46
Christopher.L
7571317
edited Nov 22 at 21:34
user26857
39.2k123882
asked Nov 22 at 15:46
Christopher.L
7571317
edited Nov 22 at 21:34
user26857
39.2k123882
edited Nov 22 at 21:34
user26857
39.2k123882
edited Nov 22 at 21:34
user26857
39.2k123882
39.2k123882
asked Nov 22 at 15:46
Christopher.L
7571317
asked Nov 22 at 15:46
Christopher.L
7571317
asked Nov 22 at 15:46
Christopher.L
7571317
7571317
add a comment |
add a comment |
1 Answer
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oldest
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1
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The answer to your first question is that $K$ is not necessarily a submodule of $N,$ in which case the expression $N/K$ doesn't make sense! $N + K$ is the smallest submodule of $M$ containing both $N$ and $K,$ so if you want to "take a quotient of $N$ by $K,$" then what you really want is $(N+K)/K.$ If $K$ is already a submodule of $N,$ then $N + K = N$ and no harm is done.
Proof that $varphi(N)= (N+K)/K$: (you've basically done this already - this is essentially the argument you wanted to make to show that $"N/K" = (N+K)/K$!)
For $min M,$ $varphi(m) = m+K.$ Thus, $varphi(N) = {n + Kmid nin Nsubseteq M}.$ You've already noted that $(N + K)/K = {(n + k) + Kmid nin N, kin K},$ and that for any $(n + k) + Kin (N+K)/K,$ we have $(n + k) + K = (n + 0) + K = n + K.$ Thus, each element of $(N + K)/K$ has a representative in $N$ (that each element of $N$ represents an element of $(N+K)/K$ is trivially true), and so $varphi(N) = (N+K)/K.$
Essentially, the argument you've made just shows that the module map
begin{align*}
f : varphi(N)&to (N+K)/K\
n + K&mapsto (n + 0) + K,
end{align*}
is an isomorphism. Upon examination, one also sees that $f$ is in fact the identity, considering $varphi(N)$ and $(N+K)/K$ inside $M/K.$
answered Nov 22 at 20:27
Stahl
16.2k43354
1
Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
– Christopher.L
Nov 23 at 13:21
add a comment |
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1 Answer
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1 Answer
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active
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accepted
The answer to your first question is that $K$ is not necessarily a submodule of $N,$ in which case the expression $N/K$ doesn't make sense! $N + K$ is the smallest submodule of $M$ containing both $N$ and $K,$ so if you want to "take a quotient of $N$ by $K,$" then what you really want is $(N+K)/K.$ If $K$ is already a submodule of $N,$ then $N + K = N$ and no harm is done.
Proof that $varphi(N)= (N+K)/K$: (you've basically done this already - this is essentially the argument you wanted to make to show that $"N/K" = (N+K)/K$!)
For $min M,$ $varphi(m) = m+K.$ Thus, $varphi(N) = {n + Kmid nin Nsubseteq M}.$ You've already noted that $(N + K)/K = {(n + k) + Kmid nin N, kin K},$ and that for any $(n + k) + Kin (N+K)/K,$ we have $(n + k) + K = (n + 0) + K = n + K.$ Thus, each element of $(N + K)/K$ has a representative in $N$ (that each element of $N$ represents an element of $(N+K)/K$ is trivially true), and so $varphi(N) = (N+K)/K.$
Essentially, the argument you've made just shows that the module map
begin{align*}
f : varphi(N)&to (N+K)/K\
n + K&mapsto (n + 0) + K,
end{align*}
is an isomorphism. Upon examination, one also sees that $f$ is in fact the identity, considering $varphi(N)$ and $(N+K)/K$ inside $M/K.$
answered Nov 22 at 20:27
Stahl
16.2k43354
1
Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
– Christopher.L
Nov 23 at 13:21
add a comment |
up vote
1
down vote
accepted
The answer to your first question is that $K$ is not necessarily a submodule of $N,$ in which case the expression $N/K$ doesn't make sense! $N + K$ is the smallest submodule of $M$ containing both $N$ and $K,$ so if you want to "take a quotient of $N$ by $K,$" then what you really want is $(N+K)/K.$ If $K$ is already a submodule of $N,$ then $N + K = N$ and no harm is done.
Proof that $varphi(N)= (N+K)/K$: (you've basically done this already - this is essentially the argument you wanted to make to show that $"N/K" = (N+K)/K$!)
For $min M,$ $varphi(m) = m+K.$ Thus, $varphi(N) = {n + Kmid nin Nsubseteq M}.$ You've already noted that $(N + K)/K = {(n + k) + Kmid nin N, kin K},$ and that for any $(n + k) + Kin (N+K)/K,$ we have $(n + k) + K = (n + 0) + K = n + K.$ Thus, each element of $(N + K)/K$ has a representative in $N$ (that each element of $N$ represents an element of $(N+K)/K$ is trivially true), and so $varphi(N) = (N+K)/K.$
Essentially, the argument you've made just shows that the module map
begin{align*}
f : varphi(N)&to (N+K)/K\
n + K&mapsto (n + 0) + K,
end{align*}
is an isomorphism. Upon examination, one also sees that $f$ is in fact the identity, considering $varphi(N)$ and $(N+K)/K$ inside $M/K.$
answered Nov 22 at 20:27
Stahl
16.2k43354
1
Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
– Christopher.L
Nov 23 at 13:21
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The answer to your first question is that $K$ is not necessarily a submodule of $N,$ in which case the expression $N/K$ doesn't make sense! $N + K$ is the smallest submodule of $M$ containing both $N$ and $K,$ so if you want to "take a quotient of $N$ by $K,$" then what you really want is $(N+K)/K.$ If $K$ is already a submodule of $N,$ then $N + K = N$ and no harm is done.
Proof that $varphi(N)= (N+K)/K$: (you've basically done this already - this is essentially the argument you wanted to make to show that $"N/K" = (N+K)/K$!)
For $min M,$ $varphi(m) = m+K.$ Thus, $varphi(N) = {n + Kmid nin Nsubseteq M}.$ You've already noted that $(N + K)/K = {(n + k) + Kmid nin N, kin K},$ and that for any $(n + k) + Kin (N+K)/K,$ we have $(n + k) + K = (n + 0) + K = n + K.$ Thus, each element of $(N + K)/K$ has a representative in $N$ (that each element of $N$ represents an element of $(N+K)/K$ is trivially true), and so $varphi(N) = (N+K)/K.$
Essentially, the argument you've made just shows that the module map
begin{align*}
f : varphi(N)&to (N+K)/K\
n + K&mapsto (n + 0) + K,
end{align*}
is an isomorphism. Upon examination, one also sees that $f$ is in fact the identity, considering $varphi(N)$ and $(N+K)/K$ inside $M/K.$
answered Nov 22 at 20:27
Stahl
16.2k43354
The answer to your first question is that $K$ is not necessarily a submodule of $N,$ in which case the expression $N/K$ doesn't make sense! $N + K$ is the smallest submodule of $M$ containing both $N$ and $K,$ so if you want to "take a quotient of $N$ by $K,$" then what you really want is $(N+K)/K.$ If $K$ is already a submodule of $N,$ then $N + K = N$ and no harm is done.
Proof that $varphi(N)= (N+K)/K$: (you've basically done this already - this is essentially the argument you wanted to make to show that $"N/K" = (N+K)/K$!)
For $min M,$ $varphi(m) = m+K.$ Thus, $varphi(N) = {n + Kmid nin Nsubseteq M}.$ You've already noted that $(N + K)/K = {(n + k) + Kmid nin N, kin K},$ and that for any $(n + k) + Kin (N+K)/K,$ we have $(n + k) + K = (n + 0) + K = n + K.$ Thus, each element of $(N + K)/K$ has a representative in $N$ (that each element of $N$ represents an element of $(N+K)/K$ is trivially true), and so $varphi(N) = (N+K)/K.$
Essentially, the argument you've made just shows that the module map
begin{align*}
f : varphi(N)&to (N+K)/K\
n + K&mapsto (n + 0) + K,
end{align*}
is an isomorphism. Upon examination, one also sees that $f$ is in fact the identity, considering $varphi(N)$ and $(N+K)/K$ inside $M/K.$
answered Nov 22 at 20:27
Stahl
16.2k43354
edited Nov 22 at 20:50
answered Nov 22 at 20:27
Stahl
16.2k43354
answered Nov 22 at 20:27
Stahl
16.2k43354
answered Nov 22 at 20:27
Stahl
16.2k43354
16.2k43354
1
Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
– Christopher.L
Nov 23 at 13:21
add a comment |
1
Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
– Christopher.L
Nov 23 at 13:21
1
1
Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
– Christopher.L
Nov 23 at 13:21
Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
– Christopher.L
Nov 23 at 13:21
add a comment |
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